\(A=2+2^2+2^3+...+2^{2023}\)

\(2A=2^2+2^3+2^4+...+2^{2024}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2024}\right)-\left(2+2^2+2^3+...+2^{2023}\right)\)

\(A=2^{2024}-2\)

\(B=3^0+3^1+3^2+...+3^{100}\)

\(3B=3^1+3^2+3^3+...+3^{101}\)

\(3B-B=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(3^0+3^1+3^2+...+3^{100}\right)\)

\(2B=3^{101}-3\)

\(B=\dfrac{3^{101}-3}{2}\)

\(C=4^0+4^2+...+4^{100}\)

=>\(16C=4^2+4^4+...+4^{102}\)

=>\(16C-C=4^2+4^4+...+4^{102}-4^0-4^2-...-4^{100}\)

=>\(15C=4^{102}-1\)

=>\(C=\dfrac{4^{102}-1}{15}\)

\(D=1+5^2+5^4+...+5^{2022}\)

=>\(25D=5^2+5^4+...+5^{2024}\)

=>\(25D-D=5^2+5^4+...+5^{2024}-1-5^2-...-5^{2022}\)

=>\(24D=5^{2024}-1\)

=>\(D=\dfrac{5^{2024}-1}{24}\)

\(S=a^0+a^1+...+a^n\)

=>\(S\cdot a=a^1+a^2+...+a^{n+1}\)

=>\(S\cdot a-S=a^1+a^2+...+a^{n+1}-a^0-a^1-...-a^n\)

=>\(S\left(a-1\right)=a^{n+1}-1\)

=>\(S=\dfrac{a^{n+1}-1}{a-1}\)

`Ư(16) = {-16;-8;-4;-2;-1;1;2;4;8;16}`

`Ư(24) = {-24;-12;-8;-6;-4;-3;-2;-1;1;2;3;4;6;8;12;24}`

`=> ƯC(16;24) = {-8;-4;-2;-1;1;2;4;8}`

`Ư(30) = {-30;-15;-10;-6;-5;-3-2;-1;1;2;3;5;6;10;15;30}`

`Ư(45) = {-45;-15;-9;-5;-3;-1;1;3;5;9;15;45}`

`=> ƯC(30;45) = {-15;-5;-3;-1;1;3;5;15}`

Lời giải:

a. $140=2^2.5.7$

$168=2^3.3.7$

$\Rightarrow ƯCLN(140, 168) = 2^2.7=28$

b. 

$525=3.5^2.7$

$375=3.5^3$

$\Rightarrow ƯCLN(525, 375)=3.5^2=75$

Do `35 vdots x; 105 vdots x`

`=> x in UC{35;105)`

Mà `105 vdots 35`

`=> x in Ư(35) = {1;5;7;35}`

Mà `x > 5 -> x in {7;35}`

Vậy ...

`b) x vdots 10; x vdots 15`

`=> x in BC(10;15)`

Ta có: 

`10 = 2 . 5`

`15 = 3.5`

`=> BCN``N(10;15) = 2.3.5 = 30`

`=> x in B(30) = {0;30;60;90;120;...}`

Mà `x < 100 -> x in {0;30;60;90}`

a) Ta có: 

`84 = 2^2 . 3 . 7`

`108 = 2^2 . 3^3`

`=> UCLN(84;108) = 2^2 . 3 = 12`

`=> UC(84;108) = Ư(12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12}`

`x` \(\in B\left(11\right),10< x< 40\)

\(\Rightarrow B\left(11\right)=\left\{0;11;22;33;44;...\right\}\)

Mà \(10< x< 40\)

\(\Rightarrow x\in\left\{11;22;33\right\}\)

x = 11;22;33

`326 + (153 - x) = 403`

`=> 153 - x = 403 - 326`

`=> 153-x=77`

`=> x=153-77`

`=>x=76`

Vậy: `x=76`

Ta có: `3311 = 11 .301`

`->` Số `3311` là hợp số.

hợp số bạn nhé

Vì 3311 = 11 . 301 nên 3311 có ước là 11 và 301. Vậy 3311 là một hợp số.

25 = 5²

Ư(25) = {1; 5; 25}